how do you do the following integral :
int (4t / (1+6t^2))
I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.

Question

how do you do the following integral : 
 int (4t / (1+6t^2)) 
I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.

anonymous · Answer

$$\int \frac{4t}{1+6t^2}$$

anonymous · Answer

$$4\int \frac{t}{1+6t^2}dt $$
$$u=1+6t^2$$
$$du=12tdt$$
$$\frac{du}{12}=tdt$$

anonymous · Answer

Ah you used substitution. Thanks a lot, makes muh more sense

anonymous · Answer

ys and it's 1/3

anonymous · Answer

wolframalpha is a good site to use to check

campbell_st · Answer

ok... so it looks like a log function to start

s = 1 + 6u^2    ds/dt = 12t

but you only need 4t   so it becomes 1/3 of ds/dt

so 

$$\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt$$

anonymous · Answer

the 1/8 might be a typo

anonymous · Answer

Thanks to both of you :) help is greatly appreciated haha. And it looks like it. 1/3 does seem right!